Asked by Anonymous
A small 5.00kg brick is released from rest 2.00m above a horizontal seesaw on a fulcrum at its center, as shown in the figure below . (radius of horizontal seesaw is 1.6m)
a.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.
b.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.
a.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.
b.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.
Answers
Answered by
anoymous student
the answer to part a) is zero. I don't know why right now though. and the answer to part b) would be 45.1 if it was a 4.50kg brick. I also don't remember how I got that answer.
Answered by
anon
For part B:
angular momentum =
m*r*sqrt(2gh), where r=1.6 and h=2
5*1.6*sqrt(2*9.8*2)
angular momentum =
m*r*sqrt(2gh), where r=1.6 and h=2
5*1.6*sqrt(2*9.8*2)
Answered by
AmyZ
Part 1. Velocity is 0.
(mass)(velocity)(r) = 0
Part 2:
v= (sqrt (2gh))= (sqrt(2(9.8)(2))=6.26
(mass)(velocity)(r)
(4.5)(6.26)(1.6)= 45.072=45.1
(mass)(velocity)(r) = 0
Part 2:
v= (sqrt (2gh))= (sqrt(2(9.8)(2))=6.26
(mass)(velocity)(r)
(4.5)(6.26)(1.6)= 45.072=45.1
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